Proof:

Let $A$ be some matrix, we want to prove that every matrix is row equivalent to exactly one reduced echelon matrix.

Let us assume that a matrix $A$ is row equivalent to more than one reduced echelon matrix. Specifically, suppose there exist two reduced echelon matrices $B$ and $C$ such that both are row equivalent to $A$.

Since both $B$ and $C$ are row equivalent to $A$, there exist sequences of row operations that transform $A$ into $B$ and $A$ into $C$.

Since row operations preserve the pivot positions, the pivot positions in $B$ must be the same as those in $A$. Similarly, the pivot positions in $C$ must also match those in $A$.

Since both $B$ and $C$ are in reduced echelon form and have the same pivot positions, they must be identical. This contradicts our assumption that $B$ and $C$ are distinct reduced echelon matrices. Therefore, the assumption that a matrix can be row equivalent to more than one reduced echelon matrix is false. It follows that every matrix is row equivalent to exactly one reduced echelon matrix.

Important:

For this proof, since it is an "if and only if," we need to prove both the forward phase (denoted as $\Rightarrow$) and the backward phase (denoted as $\Leftarrow$) of the theorem.

Lets first do the forward phase of the proof: A linear system is consistent if the rightmost column of the augmented matrix is not a pivot column.

Proof $\Rightarrow$ (By contradiction):

Lets assume that a linear system is consistent if the rightmost column of the augmented matrix is a pivot column. Now, let $A$ be the augmented matrix for this system, and let $A$ be in reduced echelon form. We can denote the columns of $A$ as $[x_{1} x_{2} x_{3} \dots x_n \mid b]$, where $b$ is the augmented column. Recall that all values above and below a pivot position are zeroed out, and since we have a pivot position in our augmented column, we have a row in the form $[0\, 0\, \dots\, 0 \mid b]$, where $b$ is the pivot position.

Looking at this system as an equation, we get $0x_{1} + 0x_{2} + 0x_{3} + \dots + 0x_{n} = b$, where $b$ is non-zero (by the definition of pivot position). Since $b$ is non-zero, we have $0 + 0 + 0 + \dots + 0 = b$, which is an inconsistent system since $0 + 0 + \dots + 0$ cannot equal a non-zero number. This is a contradiction to our original assumption that a linear system is consistent if the rightmost column of the augmented matrix is a pivot column.

Now, lets do the backward phase: If the rightmost column of the augmented matrix is not a pivot column, then the linear system is consistent.

Proof $\Leftarrow$ (By contradiction):

Assume that if the rightmost column of the augmented matrix is not a pivot column, then the linear system is inconsistent. In reduced echelon form, this implies that there will never be a row in the form $[0\, 0\, \dots\, 0 \mid b]$ with $b \neq 0$, because we originally stated that the rightmost column of the augmented matrix is not a pivot column. Since no contradictions arise from this, the system must be consistent. Therefore, if the rightmost column is not a pivot column, then the system is consistent.

Proof: